# Limits

• Nov 28th 2009, 10:07 AM
sderosa518
Limits
Assuming the result lim x approaches 0 (sinx)/(x) = 1, evalute

1) lim x approaches 0 (1-cosx)/(x)

2) lim x approaches 0 (1-cosx)(x^2)

3) lim x approaches 0 (tanx-sinx)/(x^3)

Can anyone help, I would appreciate it. My trig is horrible.
• Nov 28th 2009, 10:27 AM
Quote:

Originally Posted by sderosa518
Assuming the result lim x approaches 0 (sinx)/(x) = 1, evalute

1) lim x approaches 0 (1-cosx)/(x)

2) lim x approaches 0 (1-cosx)(x^2)

3) lim x approaches 0 (tanx-sinx)/(x^3)

Can anyone help, I would appreciate it. My trig is horrible.

Remember that you can use l-hopitals rule for indeterminate forms. Have you tried that?

Here's an example, number 1 should be easy enough for you if you apply l-hopitals rule, here is the second limit:

$\lim_{x->0}\frac{1-cos(x)}{x^2}=\lim_{x->0}\frac{sin(x)}{2x}$

$=\frac{1}{2}\lim_{x->0}\frac{sin(x)}{x}=\frac{1}{2}$

As you can see, there isn't much trig involved in that one. It's just l-hopitals rule.
• Nov 28th 2009, 10:34 AM
Paperwings
You can use L'Hopital's rule. For (2) You can use the property of

$\lim_{x\to0}\frac{sin(x)}{x} = 1$

to solve for the limit after doing L'Hopital's rule.
• Nov 28th 2009, 11:03 AM
Soroban
Hello, sderosa518!

Quote:

Assuming: . $\lim_{x\to0}\frac{\sin x}{x} \:=\:1$

evaluate: . $1)\;\;\lim_{x\to0}\frac{1-\cos x}{x}$

We have: . $\frac{1-\cos x}{x}$

Multiply by $\frac{1+\cos x}{1 + \cos x}\!:\quad\frac{1-\cos x}{x}\cdot\frac{1+\cos x}{1 + \cos x} \;=\;\frac{1-\cos^2\!x}{x(1+\cos x)} \;=\;\frac{\sin^2\!x}{x(1+\cos x)}$ . $= \;\frac{\sin x}{x}\cdot\frac{\sin x}{1 + \cos x}$

Then: . $\lim_{x\to0}\left(\frac{\sin x}{x}\right)\left(\frac{\sin x}{1 + \cos x}\right) \;=\;(1)\left(\frac{0}{1+1}\right) \;=\;0$

Quote:

$2)\;\;\lim_{x\to0}\frac{1-\cos x}{x^2}$

Multiply by $\frac{1+\cos x}{1+\cos x}\!:\quad \frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x} \;=\;\frac{1-\cos^2\!x}{x^2(1+\cos x)}$ . $=\;\frac{\sin^2\!x}{x^2(1+\cos x)} \;=\;\frac{\sin^2\!x}{x^2}\cdot\frac{1}{1+\cos x}$

Then: . $\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\left(\frac{1}{1+\cos x}\right) \;=\;(1^2)\left(\frac{1}{1+1}\right) \;=\;\frac{1}{2}$

Quote:

$3)\;\;\lim_{x\to0}\frac{\tan x - \sin x}{x^3}$

We have: . $\frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \;=\;\frac{\sin x - \sin x\cos x}{x^3\cos x} \;=\;\frac{\sin x(1 - \cos x)}{x^3\cos x}$

Multiply by $\frac{1+\cos x}{1+\cos x}\!:\quad \frac{\sin x(1 - \cos x)}{x^3\cos x}\cdot\frac{1+\cos x}{1 + \cos x} \;=\;\frac{\sin x(1-\cos^2\!x)}{x^3\cos x(1 + \cos x)}$

. . . . . . $= \; \frac{\sin x\cdot\sin^2\!x}{x^3\cos x(1 + \cos x)} \;=\;\frac{\sin^3\!x}{x^3\cos x(1 + \cos x)} \;=\;\frac{\sin^3\!x}{x^3}\cdot\frac{1}{\cos x(1 + \cos x)}$

Then: . $\lim_{x\to0}\left[\frac{\sin^3\!x}{x^3}\cdot\frac{1}{\cos x(1 + \cos x)}\right] \;=\;(1^3)\left(\frac{1}{1(1+1)}\right) \;=\;\frac{1}{2}$